∫cos4xdxsin3x(sin5x+cos5x)35=−12(1+cotAx)B+C then AB=
A
1
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B
2
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C
12
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D
None of these
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Solution
The correct option is B 2 I=∫cosx4dxsin3x(sin5x+cos5x)35=∫cosec2xcot4xdx(1+cot5x)3/51+cot5x=t.5cot4x(−cosec2x)dx=dt=−15∫dtt3/5=−t2/52=−12(1+cot5x)2/5+C.